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where m is the smallest integer such that 1^ is not isomorphic to any of the
ideals Il5 . . . , \k . (If all the 1^ are isomorphic to one or other of Il9 . . . , lk, the
induction stops at lk .) Let J be the sequence of indices k so obtained, and for
each k e J let lj be the sequence of integers n such that 1^ is isomorphic to Ifc .
We define ak to be the Hilbert sum of the 1^ for n e lk . Clearly A is the Hilbert
sum of the left ideals ak (6.4.2).

Let I be any minimal left ideal in A. Then I must be isomorphic to one of
the Ifc, for otherwise it would be orthogonal to all the \'n (1 5.8.1 2(iii)) and
hence orthogonal to A itself, which is absurd. The same argument shows that I
is orthogonal to all the ah with h ^ k. Hence, as ak is the orthogonal supple-
ment of the Hilbert sum of the ah such that h ^ k, we must have I c ak . From
this it follows already that ak is the closure of the sum of all the minimal left
ideals of A which are isomorphic to l^, and therefore ak is independent of the
decomposition of A as the Hilbert sum of the 1^, from which we started.
Moreover, for each x e A and each n e Ifc, Vnx is a left ideal which is either
{0} or isomorphic to 1^ (15.8.12(iv)), hence is contained in afc. This proves
that ak is a two-sided ideal. If 1^ = Ae'n , where e'n is an irreducible self-adjoint
idempotent, then V* = e'nA, hence a* = ak.

Let I" be a minimal left ideal of the Hilbert algebra ak . We have I" = ake",
where e" is a self-adjoint idempotent (15.8.7), and e'ne" cannot vanish for all
n e lk , otherwise I" would be orthogonal to all the 1^ (n e lk) and therefore
to the closure of their sum, namely to ak : which is absurd because I" ^ {0},
Hence there exists at least one index n E lk such that l'n\'r ^ {0}; since 1^1" is a
left ideal in ak , we must have 1^ I" = I", which shows that I" is a minimal left
ideal of A, necessarily isomorphic to Vn and therefore to \k (1 5.8.1 2(iii)). If
now b is a nonzero two-sided ideal of the algebra ak , it contains at least
one minimal left ideal I" of this algebra (15.8.8), hence also contains
all the H; (/lei*). But \"\'n = i; (1 5.8.1 2(iii)), and therefore b contains
the sum of all the \'n (n e lk). If b is closed, it follows that b = ak. Finally,
ah n ak = {0} if h ^ k, because ah and ak are two-sided ideals.

A complete Hilbert algebra A is said to be topologically simple if it con-
tains no closed two-sided ideal other than A and {0}. It follows from (15.8.13)
that the study of the structure of a separable complete Hilbert algebra A
reduces completely to the study of the afc, that is to the case where A is
topologically simple.

(1 5.8.1 4) Let Abe a topologically simple, complete, separable Hilbert algebra.
Then for each minimal left ideal I of A, the representation jcf > U^x) of A in the
Hilbert space I is faithful.

If A is infinite-dimensional, then so is I. The image of A under Ul is the
algebra J$?2(0 of Hitbert-Schmidt operators on I (15.4.8), and there exists ath a, such that b2 =